A regular tetrahedron and the methane angle

Q22 Figure1 methane and Figure2 tetrahedron QRSTHHHHCFigure 1TQRSVFigure 2
Question diagram 2
Q22 Figure1 methane and Figure2 tetrahedron QRSTHHHHCFigure 1TQRSVFigure 2

Native methane molecule and tetrahedron, Figure 1 and Figure 2

Figure 1 shows a methane molecule consisting of a carbon atom with four hydrogen atoms symmetrically placed around it. Figure 2 shows the tetrahedron structure of the methane molecule with the centres of the hydrogen atoms represented by points Q,R,S,TQ,R,S,T and the centre of the carbon atom represented by point VV.

The points Q,R,SQ,R,S have coordinates (8,1,8),(8,7,2),(2,1,2)(8,1,8),(8,7,2),(2,1,2) respectively and form an equilateral triangle.

(i) Find a cartesian equation of the plane pp which passes through the points Q,R,SQ,R,S. marks

(ii) Find a cartesian equation of the plane p1p_1 which passes through the midpoint of QRQR and is perpendicular to QRQR. marks

Plane p2p_2 which passes through the midpoint of RSRS and is perpendicular to RSRS has equation x+y=9x+y=9.

(iii) Find a vector equation of the line ll where p1,p2p_1,p_2 meet. mark

(iv) The point TT is on the line ll such that QRSTQRST is a regular tetrahedron with QR=QTQR=QT. Show that the possible coordinates for point TT is (2,7,8)(2,7,8). Hence, or otherwise, find the coordinates of a point on plane pp that is closest to point TT. marks

(v) Given that TVTV is 34\frac34 of the vertical height of tetrahedron QRSTQRST. Find the coordinates of point VV and hence show that the bonding angle TVQTVQ of the methane molecule is 109.5∘109.5^\circ (correct to 1 decimal place). marks

[2018/ACJC/Prelim/I/11]

Total marks: 16

Solution

(i)

Plane through three points

Q(8,1,8)Q(8,1,8), R(8,7,2)R(8,7,2), S(2,1,2)S(2,1,2) form an equilateral triangle. Find a Cartesian equation of the plane pp through them.

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Hint

  1. Normal =QR→×QS→=\overrightarrow{QR}\times\overrightarrow{QS}.

Answer · (i)

p:−x+y+z=1p:-x+y+z=1.

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The base triangle QRS

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(ii)

Perpendicular bisector plane

Find the plane p1p_1 through the midpoint of QRQR perpendicular to QRQR.

Hint

  1. Normal =QR→∥(0,1,−1)=\overrightarrow{QR}\parallel(0,1,-1).

Answer · (ii)

p1:y−z=−1p_1:y-z=-1.

(iii)

Line of intersection

p2:x+y=9p_2:x+y=9. Find a vector equation of l=p1∩p2l=p_1\cap p_2.

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Hint

  1. Solve y−z=−1y-z=-1 and x+y=9x+y=9.

Answer · (iii)

r=(10,−1,0)+t(−1,1,1)r=(10,-1,0)+t(-1,1,1).

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The line through the apexes

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(iv)

The fourth vertex

T∈lT\in l with QT=QRQT=QR. Show T=(2,7,8)T=(2,7,8) is a possible position and find the point of pp closest to TT.

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Hint

  1. T=(10−λ,−1+λ,λ)T=(10-\lambda,-1+\lambda,\lambda).

  2. ∣QT∣2=∣QR∣2=72|QT|^2=|QR|^2=72.

Answer · (iv)

One vertex T=(2,7,8)T=(2,7,8); foot (6,3,4)(6,3,4).

The other tetrahedron vertex is (10,−1,0)(10,-1,0).

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The regular tetrahedron QRST

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(v)

The methane bond angle

VV divides the height so that TV=34TV=\frac34 of it. Find VV and show ∠TVQ=109.5∘\angle TVQ=109.5^\circ.

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Hint

  1. V=3F+T4V=\frac{3F+T}4.

  2. Angle between VQ→\overrightarrow{VQ} and VT→\overrightarrow{VT}.

Answer · (v)

V=(5,4,5)V=(5,4,5); ∠TVQ=cos⁡−1(−1/3)=109.5∘\angle TVQ=\cos^{-1}(-1/3)=109.5^\circ.

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The carbon atom at the centre

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